A Subexponential Parameterized Algorithm for Proper Interval Completion

A subexponential parameterized algorithm for Proper Interval Completion

Ivan Bliznets1 Fedor V. Fomin2 Marcin Pilipczuk2 Micha l Pilipczuk2

1St. Petersburg Academic University of the Russian Academy of Sciences, Russia

2Department of Informatics, University of Bergen, Norway

ESA’14, Wroc law, September 9th, 2014

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 1/17 Proper interval graphs: graphs admitting an intersection model of intervals on a line s.t. no interval contains any other interval. Unit interval graphs: graphs admitting an intersection model of unit intervals on a line. PIG = UIG.

(Proper) interval graphs

Interval graphs: graphs admitting an intersection model of intervals on a line.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 2/17 Unit interval graphs: graphs admitting an intersection model of unit intervals on a line. PIG = UIG.

(Proper) interval graphs

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 2/17 PIG = UIG.

(Proper) interval graphs

Interval graphs: graphs admitting an intersection model of intervals on a line. Proper interval graphs: graphs admitting an intersection model of intervals on a line s.t. no interval contains any other interval. Unit interval graphs: graphs admitting an intersection model of unit intervals on a line.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 2/17 (Proper) interval graphs

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 2/17 This talk: FPT algorithms for PIC (time f (k) · nO(1)) Related paper: the same about IC

The problem

(Proper) Interval Completion Input: A graph G and an integer k Parameter: k Question: Can one turn G into a (proper) interval graph by adding at most k edges?

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 3/17 Related paper: the same about IC

The problem

(Proper) Interval Completion Input: A graph G and an integer k Parameter: k Question: Can one turn G into a (proper) interval graph by adding at most k edges?

This talk: FPT algorithms for PIC (time f (k) · nO(1))

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 3/17 The problem

(Proper) Interval Completion Input: A graph G and an integer k Parameter: k Question: Can one turn G into a (proper) interval graph by adding at most k edges?

This talk: FPT algorithms for PIC (time f (k) · nO(1)) Related paper: the same about IC

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 3/17 Approach via deleting forbidden induced subgraphs. For PIC, the running time was later reduced to 4k · nO(1). Villanger et al., 2007: a k2k · nO(1) algorithm for IC. Fomin and Villanger, 2012: a kO(k1/2) · nO(1) algorithm for Chordal Completion. Do other completion problems admit subexponential parameterized algorithms?

History

Kaplan et al., 1994: 16k · nO(1) algorithms for completion to chordal and proper interval graphs.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 4/17 For PIC, the running time was later reduced to 4k · nO(1). Villanger et al., 2007: a k2k · nO(1) algorithm for IC. Fomin and Villanger, 2012: a kO(k1/2) · nO(1) algorithm for Chordal Completion. Do other completion problems admit subexponential parameterized algorithms?

History

Kaplan et al., 1994: 16k · nO(1) algorithms for completion to chordal and proper interval graphs. Approach via deleting forbidden induced subgraphs.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 4/17 Villanger et al., 2007: a k2k · nO(1) algorithm for IC. Fomin and Villanger, 2012: a kO(k1/2) · nO(1) algorithm for Chordal Completion. Do other completion problems admit subexponential parameterized algorithms?

History

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 4/17 Fomin and Villanger, 2012: a kO(k1/2) · nO(1) algorithm for Chordal Completion. Do other completion problems admit subexponential parameterized algorithms?

History

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 4/17 Do other completion problems admit subexponential parameterized algorithms?

History

Kaplan et al., 1994: 16k · nO(1) algorithms for completion to chordal and proper interval graphs. Approach via deleting forbidden induced subgraphs. For PIC, the running time was later reduced to 4k · nO(1). Villanger et al., 2007: a k2k · nO(1) algorithm for IC. Fomin and Villanger, 2012: a kO(k1/2) · nO(1) algorithm for Chordal Completion.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 4/17 History

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 4/17 GKKMPRR: 1/2 O?(kO(k ))

DFPV: DFPV: 1/2 1/2 O?(kO(k )) O?(kO(k ))

BFPP: FV: 1/2 1/2 O?(kO(k )) O?(kO(k ))

BFPP: 2/3 O?(kO(k ))

Subexponential algorithms

Split

⊂ ⊂ Threshold ⊂ Trivially perfect ⊂

Interval ⊂ Chordal

⊂ Proper interval

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 5/17 GKKMPRR: 1/2 O?(kO(k ))

DFPV: DFPV: 1/2 1/2 O?(kO(k )) O?(kO(k ))

BFPP: 1/2 O?(kO(k ))

BFPP: 2/3 O?(kO(k ))

Subexponential algorithms

Split

⊂ ⊂ Threshold ⊂ Trivially perfect ⊂

Interval ⊂ Chordal

⊂ FV: ? O(k1/2) Proper interval O (k )

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 5/17 DFPV: DFPV: 1/2 1/2 O?(kO(k )) O?(kO(k ))

BFPP: 1/2 O?(kO(k ))

BFPP: 2/3 O?(kO(k ))

Subexponential algorithms

GKKMPRR: 1/2 O?(kO(k )) Split

⊂ ⊂ Threshold ⊂ Trivially perfect ⊂

Interval ⊂ Chordal

⊂ FV: ? O(k1/2) Proper interval O (k )

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 5/17 BFPP: 1/2 O?(kO(k ))

BFPP: 2/3 O?(kO(k ))

Subexponential algorithms

GKKMPRR: 1/2 O?(kO(k )) Split

⊂ ⊂ Threshold ⊂ Trivially perfect DFPV: DFPV: ⊂ ? O(k1/2) ? O(k1/2) O (k ) O (k ) Interval ⊂ Chordal

⊂ FV: ? O(k1/2) Proper interval O (k )

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 5/17 BFPP: 2/3 O?(kO(k ))

Subexponential algorithms

GKKMPRR: 1/2 O?(kO(k )) Split

⊂ ⊂ Threshold ⊂ Trivially perfect DFPV: DFPV: ⊂ ? O(k1/2) ? O(k1/2) O (k ) O (k ) Interval ⊂ Chordal

⊂ BFPP: FV: ? O(k1/2) ? O(k1/2) Proper interval O (k ) O (k )

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 5/17 Subexponential algorithms

GKKMPRR: 1/2 O?(kO(k )) Split

⊂ ⊂ Threshold ⊂ Trivially perfect DFPV: DFPV: ⊂ ? O(k1/2) ? O(k1/2) O (k ) O (k ) Interval ⊂ Chordal

⊂ BFPP: FV: ? O(k1/2) ? O(k1/2) Proper interval O (k ) O (k ) BFPP: 2/3 O?(kO(k ))

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 5/17 Examples: C4-free Deletion, C4-free Completion, Trivially Perfect Deletion, Cograph Completion... (DFPV). Essentially, the presented completion problems are singular cases for which subexponential parameterized algorithms are possible.

SUBEPT and ETH

For many other edge modiﬁcation problems, under ETH one can exclude existence of a 2o(k) · nO(1) algorithm.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 6/17 Essentially, the presented completion problems are singular cases for which subexponential parameterized algorithms are possible.

SUBEPT and ETH

For many other edge modiﬁcation problems, under ETH one can exclude existence of a 2o(k) · nO(1) algorithm.

Examples: C4-free Deletion, C4-free Completion, Trivially Perfect Deletion, Cograph Completion... (DFPV).

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 6/17 SUBEPT and ETH

For many other edge modiﬁcation problems, under ETH one can exclude existence of a 2o(k) · nO(1) algorithm.

Examples: C4-free Deletion, C4-free Completion, Trivially Perfect Deletion, Cograph Completion... (DFPV). Essentially, the presented completion problems are singular cases for which subexponential parameterized algorithms are possible.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 6/17 Alternative view: Find a shape of the decomposition of the completed graph that requires the least number of ﬁll edges. Each of the considered graph classes has a decomposition: a clique tree, an interval model, etc. Decomposition has building blocks, e.g., maximal cliques. A chordal graph has ≤ n + 1 maximal cliques. Idea: A graph that lacks k edges to being a chordal graph, has ≤ kO(k1/2) · nO(1) sets that can become a clique after completion (potential maximal cliques).

The approach of FV

Standard view: Kill all the forbidden subgraphs by adding as few edges as possible.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 7/17 Each of the considered graph classes has a decomposition: a clique tree, an interval model, etc. Decomposition has building blocks, e.g., maximal cliques. A chordal graph has ≤ n + 1 maximal cliques. Idea: A graph that lacks k edges to being a chordal graph, has ≤ kO(k1/2) · nO(1) sets that can become a clique after completion (potential maximal cliques).

The approach of FV

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 7/17 Decomposition has building blocks, e.g., maximal cliques. A chordal graph has ≤ n + 1 maximal cliques. Idea: A graph that lacks k edges to being a chordal graph, has ≤ kO(k1/2) · nO(1) sets that can become a clique after completion (potential maximal cliques).

The approach of FV

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 7/17 A chordal graph has ≤ n + 1 maximal cliques. Idea: A graph that lacks k edges to being a chordal graph, has ≤ kO(k1/2) · nO(1) sets that can become a clique after completion (potential maximal cliques).

The approach of FV

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 7/17 Idea: A graph that lacks k edges to being a chordal graph, has ≤ kO(k1/2) · nO(1) sets that can become a clique after completion (potential maximal cliques).

The approach of FV

Standard view: Kill all the forbidden subgraphs by adding as few edges as possible. Alternative view: Find a shape of the decomposition of the completed graph that requires the least number of ﬁll edges. Each of the considered graph classes has a decomposition: a clique tree, an interval model, etc. Decomposition has building blocks, e.g., maximal cliques. A chordal graph has ≤ n + 1 maximal cliques.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 7/17 The approach of FV

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 7/17 Final DP that assembles the decomposition. Enumeration of building blocks.

Enumerate potential building blocks. Hopefully there is 2o(k) · nO(1) of them. Perform a dynamic programming algorithm that assembles the decomposition. In the DP, keep track of the minimum possible number of ﬁll edges. Ingredients:

The approach of FV

Algorithm:

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 8/17 Final DP that assembles the decomposition. Enumeration of building blocks.

Perform a dynamic programming algorithm that assembles the decomposition. In the DP, keep track of the minimum possible number of ﬁll edges. Ingredients:

The approach of FV

Algorithm: Enumerate potential building blocks. Hopefully there is 2o(k) · nO(1) of them.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 8/17 Final DP that assembles the decomposition. Enumeration of building blocks.

In the DP, keep track of the minimum possible number of ﬁll edges. Ingredients:

The approach of FV

Algorithm: Enumerate potential building blocks. Hopefully there is 2o(k) · nO(1) of them. Perform a dynamic programming algorithm that assembles the decomposition.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 8/17 Final DP that assembles the decomposition. Enumeration of building blocks.

Ingredients:

The approach of FV

Algorithm: Enumerate potential building blocks. Hopefully there is 2o(k) · nO(1) of them. Perform a dynamic programming algorithm that assembles the decomposition. In the DP, keep track of the minimum possible number of ﬁll edges.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 8/17 Final DP that assembles the decomposition. Enumeration of building blocks.

The approach of FV

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 8/17 Enumeration of building blocks.

The approach of FV

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 8/17 The approach of FV

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 8/17 Ω is a potential section; L is a subset of cc of G − Ω that will go to the left.

DP state: A pair (L, Ω), where

DP[L, Ω] is the minimum number of ﬁll edges needed to make G[L ∪ Ω] interval with Ω being the end-clique. If we had a family N capturing all the relevant states, then the DP computes the optimum solution in time poly(|N |).

First try for Interval Completion

Section: a set of intervals pinpointed by a vertical line.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 9/17 Ω is a potential section; L is a subset of cc of G − Ω that will go to the left. DP[L, Ω] is the minimum number of ﬁll edges needed to make G[L ∪ Ω] interval with Ω being the end-clique. If we had a family N capturing all the relevant states, then the DP computes the optimum solution in time poly(|N |).

First try for Interval Completion

Section: a set of intervals pinpointed by a vertical line. DP state: A pair (L, Ω), where

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 9/17 L is a subset of cc of G − Ω that will go to the left. DP[L, Ω] is the minimum number of ﬁll edges needed to make G[L ∪ Ω] interval with Ω being the end-clique. If we had a family N capturing all the relevant states, then the DP computes the optimum solution in time poly(|N |).

First try for Interval Completion

Section: a set of intervals pinpointed by a vertical line. DP state: A pair (L, Ω), where Ω is a potential section;

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 9/17 DP[L, Ω] is the minimum number of ﬁll edges needed to make G[L ∪ Ω] interval with Ω being the end-clique. If we had a family N capturing all the relevant states, then the DP computes the optimum solution in time poly(|N |).

First try for Interval Completion

Section: a set of intervals pinpointed by a vertical line. DP state: A pair (L, Ω), where Ω is a potential section; L is a subset of cc of G − Ω that will go to the left.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 9/17 If we had a family N capturing all the relevant states, then the DP computes the optimum solution in time poly(|N |).

First try for Interval Completion

Section: a set of intervals pinpointed by a vertical line. DP state: A pair (L, Ω), where Ω is a potential section; L is a subset of cc of G − Ω that will go to the left. DP[L, Ω] is the minimum number of ﬁll edges needed to make G[L ∪ Ω] interval with Ω being the end-clique.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 9/17 First try for Interval Completion

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 9/17 During construction we need to make sure intervals are closed in the same order they were opened.

Problem: The section itself is not enough!

Proposition for a state: (L, Ω, σΩ), where σΩ is an ordering of Ω. Having all possible orderings is too expensive.

First try for Proper Interval Completion

Let’s try the same approach for the states.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 10/17 During construction we need to make sure intervals are closed in the same order they were opened.

Proposition for a state: (L, Ω, σΩ), where σΩ is an ordering of Ω. Having all possible orderings is too expensive.

First try for Proper Interval Completion

Let’s try the same approach for the states. Problem: The section itself is not enough!

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 10/17 Proposition for a state: (L, Ω, σΩ), where σΩ is an ordering of Ω. Having all possible orderings is too expensive.

First try for Proper Interval Completion

Let’s try the same approach for the states. Problem: The section itself is not enough! During construction we need to make sure intervals are closed in the same order they were opened.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 10/17 Having all possible orderings is too expensive.

First try for Proper Interval Completion

Let’s try the same approach for the states. Problem: The section itself is not enough! During construction we need to make sure intervals are closed in the same order they were opened.

Proposition for a state: (L, Ω, σΩ), where σΩ is an ordering of Ω.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 10/17 First try for Proper Interval Completion

Let’s try the same approach for the states. Problem: The section itself is not enough! During construction we need to make sure intervals are closed in the same order they were opened.

Proposition for a state: (L, Ω, σΩ), where σΩ is an ordering of Ω. Having all possible orderings is too expensive.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 10/17 Bessy, Perez: O(k3) kernel for the problem, so n = O(k3). Finding 2o(k) candidates for sections is not that diﬃcult. Getting partition into left/right from a candidate is very easy. We are not able to enumerate 2o(k) candidates for the ordering.

√ O( k) Finding n √ candidates for sections is not that diﬃcult. O( k) O(1) Finding k · n √ candidates for sections is more diﬃcult. We cannot have kO( k) · nO(1) partitions into left and right! We need to remodel the whole DP to construct this partition along the way. Proper Interval Completion:

And how this really works...

Interval Completion:

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 11/17 Bessy, Perez: O(k3) kernel for the problem, so n = O(k3). Finding 2o(k) candidates for sections is not that diﬃcult. Getting partition into left/right from a candidate is very easy. We are not able to enumerate 2o(k) candidates for the ordering.

√ O( k) O(1) Finding k · n √ candidates for sections is more diﬃcult. We cannot have kO( k) · nO(1) partitions into left and right! We need to remodel the whole DP to construct this partition along the way. Proper Interval Completion:

And how this really works...

: Interval Completion√ Finding nO( k) candidates for sections is not that diﬃcult.

√ We cannot have kO( k) · nO(1) partitions into left and right! We need to remodel the whole DP to construct this partition along the way. Proper Interval Completion:

And how this really works...

: Interval Completion√ O( k) Finding n √ candidates for sections is not that diﬃcult. Finding kO( k) · nO(1) candidates for sections is more diﬃcult.

We need to remodel the whole DP to construct this partition along the way. Proper Interval Completion:

And how this really works...

Proper Interval Completion:

And how this really works...

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 11/17 Finding 2o(k) candidates for sections is not that diﬃcult. Getting partition into left/right from a candidate is very easy. We are not able to enumerate 2o(k) candidates for the ordering.

And how this really works...

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 11/17 Getting partition into left/right from a candidate is very easy. We are not able to enumerate 2o(k) candidates for the ordering.

And how this really works...

: Interval Completion√ O( k) Finding n √ candidates for sections is not that difficult. O( k) O(1) Finding k · n √ candidates for sections is more difficult. We cannot have kO( k) · nO(1) partitions into left and right! We need to remodel the whole DP to construct this partition along the way. Proper Interval Completion: Bessy, Perez: O(k3) kernel for the problem, so n = O(k3). Finding 2o(k) candidates for sections is not that difficult.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 11/17 We are not able to enumerate 2o(k) candidates for the ordering.

And how this really works...

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 11/17 And how this really works...

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 11/17 There is at most 2k2/3 of them. Guess all of them and their positions. Move to a sandwich problem. Provided that expensive vertices are guessed, there is kO(τ) = kO(k1/3) candidates for sections and left/right.

Expensive vertices and sections

Expensive vertices: vertices that have more than τ = k1/3 incident ﬁll edges.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 12/17 Guess all of them and their positions. Move to a sandwich problem. Provided that expensive vertices are guessed, there is kO(τ) = kO(k1/3) candidates for sections and left/right.

Expensive vertices and sections

Expensive vertices: vertices that have more than τ = k1/3 incident ﬁll edges. There is at most 2k2/3 of them.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 12/17 Move to a sandwich problem. Provided that expensive vertices are guessed, there is kO(τ) = kO(k1/3) candidates for sections and left/right.

Expensive vertices and sections

Expensive vertices: vertices that have more than τ = k1/3 incident ﬁll edges. There is at most 2k2/3 of them. Guess all of them and their positions.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 12/17 Provided that expensive vertices are guessed, there is kO(τ) = kO(k1/3) candidates for sections and left/right.

Expensive vertices and sections

Expensive vertices: vertices that have more than τ = k1/3 incident ﬁll edges. There is at most 2k2/3 of them. Guess all of them and their positions. Move to a sandwich problem.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 12/17 Expensive vertices and sections

Expensive vertices: vertices that have more than τ = k1/3 incident ﬁll edges. There is at most 2k2/3 of them. Guess all of them and their positions. Move to a sandwich problem. Provided that expensive vertices are guessed, there is kO(τ) = kO(k1/3) candidates for sections and left/right.

Bliznets, Fomin, Pilipczuk×2 SubExp for PIC 12/17 If there were k2/3 of them, then there would be n2k2/3 guesses for such edges. Ergo, we have kO(k2/3) candidates for sections together with their orderings, provided they are cheap — incident to at most k2/3 ﬁll edges. Layer-one DP: go from a cheap section to a cheap section. We need to compute the best possible completion between two cheap sections, assuming that all the sections in between are expensive. For this Layer-two DP.